Mathematics - 2016-11-27 (page 7 of 11)

Balarka Sen

10:19

@Kaj I can draw it just fine. It'll just be an ununderstandable doodle.

Kaj Hansen

It depends on what one means by "simply cannot be drawn"

Balarka Sen

Yes :) I was just joking.

Alessandro

you can also pick a function whose graph is dense in $\mathbb{R}^2$, that should be pretty hard to draw

Astyx

Someone has done it already

Balarka Sen

It says "image not found"

Astyx

10:23

Meh nvm

Secret

a filled square is not a function on the reals, because there are more than one y value for each x

Balarka Sen

is that like a Fermat thing, where one has drawn the picture but the pixels are too small to sketch it? :P

@Alessandro I really don't know how to construct such a thing

It'd have to jump around a lot, obviously, because a y-section only consists of a point.

usukidoll

nail polish

Kaj Hansen

Is it even possible @BalarkaSen ?

Balarka Sen

I am not sure.

Alessandro

10:28

Functions $\mathbb{R}\to\mathbb{R}$ satisfying $f(x+y)=f(x)+f(y)$ are either linear or have all kind of pathological properties, in particular they have a graph which is dense in $\mathbb{R}^2$

Kaj Hansen

Started next problem set @usukidoll ?

usukidoll

nah it's like 12:30 am

Balarka Sen

@Alessandro Weird.

usukidoll

but it's a bunch of ideal proofs

prove left ideal, prove a commutative ring is an ideal, simple ideal, zero ideal

I can't think when I'm half sleepy at this hour...I'll probably do half the proofs tomorrow

and then MOnday cuz no lectures yay

Secret

@Alessandro Indeed, and those cannot be wrote down because they are all proved to exist by the AC

Alessandro

10:30

there is some axiom of choice needed to show that are discontinuous functions satisfying that functional equation though (because you need a basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$), I'm not sure if one can prove the existence of a function with a graph dense in $\mathbb{R}^2$ in ZF

Astyx

@Alessandro How would you prove this ? (that the graph is dense in $\Bbb R^2$ if it's not linear)

Kaj Hansen

Are you in....Hawaii @usukidoll ?

Alessandro

There's a proof on wiki

usukidoll

ya

Kaj Hansen

12:30 AM would put you in the middle of the Pacific ocean

Ah, that's awesome

usukidoll

10:31

mhm

but filled with expensive high priced **#)(@*#()@*()$

Kaj Hansen

I visited UH Hilo once

usukidoll

oh that's on the Big Island

Astyx

Thanks !

Kaj Hansen

mhm

usukidoll

I go to the other UH...

cuz that's the only uni with 300 level math 400 level math 500 level math and so onnnnnnn

Kaj Hansen

10:33

Do you visit the other islands often?

usukidoll

all other unis think I'm too advanced for them ;p well yeah 100 level and 200 level math is kids play

no I odn't

don't

Kaj Hansen

I can't imagine living on an island

usukidoll

opi nail polishes are cheaper on ebay

Secret

Actually this brought out a meta question:
Given a proposition that cannot be proved by anything other than the AC, can it be shown that the proposition is always non constructable, that is, we cannot find explicit examples of the proposition in question?

usukidoll

I'm not paying $10 for it... that's food

NaCl

10:35

How do you prove that $\forall z,w\in\mathbb{C},\vert z\vert<1,\vert w\vert<2:\vert z\vert+\vert w\vert<1+\vert z\vert^2\vert w\vert^2$?

Kaj Hansen

Probably. If you NEED AC, you're gonna have to be making an infinite number of choices?

You don't need AC if a definable choice function already exists, right?

Balarka Sen

I thought about it a bit. I can't construct one by hand, so I chicken out.

usukidoll

besides all of z and w in complex numbers... ragequit

Alessandro

I don't think you can construct one explicitely but I might be wrong

Kaj Hansen

(I know next to nothing about the axiomatic foundation of math)

NaCl

10:38

Original was $\vert \frac{z-w}{1-\overline{z}w}\vert<1$

But okay... I'll open a question... :(

Astyx

What you wrote before is probably not equivalent @NaCl

NaCl

hm

usukidoll

natural ac

brrrr

NaCl

$\vert\frac{z-w}{1-\overline{z}w}\vert<1\Leftrightarrow\vert\frac{z-w}{1- \overline{z} w}\vert^2<1\Leftrightarrow\frac{z-w}{1-\overline{z}w}\cdot\overline{\frac{z-w}{1‌-\overline{z}w}}$

Secret

$$|z| < 1$$
$$|w|< 2$$
$$|z|^2\in [0,1)$$
$$|w|^2\in [0,2)$$
$$|z|^2|w|^2\in [0,2)$$
$$|z|+|w|<1+2$$
$$|z|+|w|\in [0,3)$$
$$|z|+|w|<1+|z|^2|w|^2$$
?

NaCl

10:42

Upto that we're good, aren't we?

Balarka Sen

@NaCl This is wrong. Let $z = 0.5$, $w = 1$. Then $|z| + |w| = 1.5$ and $1 + |z|^2 |w|^2 = 1 + 0.25 = 1.25$.

$1.5$ is clearly not smaller than $1.25$.

Astyx

Or is it ?

NaCl

$w$ must be smaller than 1

Balarka Sen

You said $|w| < 2$.

NaCl

typo

sorry

didn't see it early enough

Secret

10:46

In that case:

$$|z| < 1$$
$$|w|< 1$$
$$|z|^2\in [0,1)$$
$$|w|^2\in [0,1)$$
$$|z|^2|w|^2\in [0,1)$$
$$|z|+|w|<1+1$$
$$|z|+|w|\in [0,2)$$
$$|z|+|w|<1+|z|^2|w|^2$$
?

nope, don't see how 1+[0,1)=[0,2)

Astyx

Anyway if you take BalarkaSen's example with $w =0.9$ it still works

Balarka Sen

Ya. It's still wrong.

Alright, I gotta go.

NaCl

$\frac{z-w}{1-\overline{z}w}\cdot\overline{\frac{z-w}{1‌-\overline{z}w}}=\frac{‌z-w}{1-\overline{z}w}\cdot \frac{\overline{z}-\overline{w}}{1-z\overline{w}}$?

Astyx

Bye @BalarkaSen

Yes @NaCl

NaCl

Then I wonder where my error was

$\frac{‌z-w}{1-\overline{z}w}\cdot \frac{\overline{z}-\overline{w}}{1-z\overline{w}}=\frac{\vert z\vert^2-z\overline{w}-w\overline{z}+\vert w\vert^2}{1-z\overline{w}-\overline{z}w+\vert z\vert^2\vert w\vert^2}<1\Leftrightarrow \vert z\vert^2-z\overline{w}-w\overline{z}+\vert w\vert^2<1-z\overline{w}-\overline{z}w+\vert z\vert^2\vert w\vert^2\Leftrightarrow\vert z\vert^2+\vert w\vert^2<1+\vert z\vert^2\vert w\vert^2$

Astyx

10:59

Now you just have to prove $(1-a)(1-b)\gt0$ for some $a, b$

NaCl

Why?

Astyx

I'll let you ponder it

NaCl

$\vert z\vert^2+\vert w\vert^2-1-\vert z\vert^2\vert w\vert^2<0$

I thought about replacing $\vert z\vert$ with $1-x_0$ and $\vert w\vert$ with $1-x_1$, where $0\le x_0<1,0\le x_1<1$, but that didn't help at all

Astyx

Factor your expression

NaCl

$(1-a)(1-b)$ is $1-b-a+ab$, mine is $(\vert z\vert^2-1)(1-\vert w\vert^2)$

Astyx

11:06

I know

NaCl

I really don't know what to do now....

Astyx

You can get the signs of each term

NaCl

hmm... $\vert z\vert^2-1<0$, since $\vert z\vert<1$, and $1-\vert w\vert^2>0$, since $\vert w\vert^2<1$

Since one factor is negative and the other one is positive, the inequality holds

And thus the whole thing is proven, isn't it?

Astyx

yup

NaCl

Thank you very much!

Astyx

11:13

my pleasure :)

Secret

11:32

How did I miss the case $z=0.5$, $w=0.9$ in my workings above?

Astyx

You didn't

Everything you wrote is fine i think

Except for the last line

Secret

So 1+[0,1)\neq [0,2)?

Or going to that line from another pathway:

$$|z|<1,|w|<1 \Rightarrow |z|+|w|<1+1$$
$$|z|+|w|-1<1$$
$$|z|+|w|-1<1$$
O wait a second...
$$|z|+|w|<0<1$$
Ah I see, the possibility of < 0 bust the pathway to reach the final line

Alessandro

12:23

@Balarka @Astyx @Kaj in this question there is a simple construction of a function with a dense graph that doesn't require any form of choice

Astyx

Thanks a lot !

NaCl

Hwo do you show that a set is not countable?

Alessandro

By showing that there is no surjection from N into your set, or no injection from your set in N, or That there is a bijection between your set and another one known to be uncountable or...

It depends on how your set was defined

NaCl

...ok

ty

Secret

I am always bad at surjection proofs and often need to rely on the following description:
$$\exists x \in S: f^\overleftarrow{}(x)=\emptyset$$

where $f^\overleftarrow{}$ is the preimage

Julian Moore

12:36

I recently asked what I thought was a straightforward question concerning the solubility of the geodesic equation. Since it has received no substantive comments and no answers, do you think it would be appropriate to ask it at mathoverflow? I didn't previously think it was a "research grade" question, but...

DHMO

12:59

Take $z=0.5$ and $w=1.5$.
Then, LHS = $0.5+1.5$ = $2$.
Also, RHS = $1+0.5^21.5^2=1.5625$.
Therefore your statement is incorrect.

@Secret ^

Never mind, I see that the question has been changed to $|z|^2+|w|^2<1+|z|^2|w|^2$

$0\le|z|<1 \land 0\le|w|<1$
$\implies 0\le|z|^2<1 \land 0\le|w|^2<1$
$\implies (|z|^2-1)<0 \land (|w|^2-1)<0$
$\implies (|z|^2-1)(|w|^2-1)>0$
$\implies |z|^2|w|^2 - |z|^2 - |w|^2 + 1 > 0$
$\implies |z|^2+|w|^2 < 1+|z|^2|w|^2$

@usukidoll question has nothing to do with complex numbers

Astyx

Wrong use of $\implies$

DHMO

@Astyx where?

Astyx

Everywhere

DHMO

why?

Astyx

$\implies$ is not associative

DHMO

13:07

implicit parentheses.

Astyx

$(A\implies B) \implies C$ is not the same as $A\implies (B\implies C)$

And neither of them are what you're trying to say

DHMO

I'm saying $A\implies B$ and $B\implies C$

Just like how you use $0<x<1$ to mean $0<x$ and $x<1$

Astyx

But that's not what you wrote

DHMO

and $0=x=1$ to mean $0=x$ and $x=1$

morbidCode

Hi, is anyone here?

DHMO

13:08

I wrote $A\implies B\implies C$, meaning $A\implies B$ and $B\implies C$

Astyx

No we're not

DHMO

@morbidCode no

Astyx

I know what you meant

I'm just saying it's not correct :)

DHMO

@Astyx well you aren't the authority

Astyx

That's true

morbidCode

13:09

Nice. Can I ask about the aproximation of pi here?

DHMO

"Correctness" is meaningless if you can understand.

> Just ask; don't ask to ask.

Astyx

I disagree, but nvm

DHMO

@Astyx people don't use $A\implies B\implies C$ to mean $A\implies B$ and $B \implies C$?

Astyx

At least they shouldn't

DHMO

who said they shouldn't?

Null

13:13

@KajHansen new picture? i hear heavy metal and all sorts of stuff :)

Astyx

The fact that it doesn't formally mean anything ?

DHMO

@Astyx Do you object to $A \iff B \iff C$ meaning $A\iff B \land B \iff C$?

Astyx

No, because $\iff$ is an equivalence relation

DHMO

well writing block of equations would be a disaster for you if you have to repeat everything twice

Astyx

That's exactly why writing blocks of equations are a disaster

DHMO

13:15

but I've never thought about that being abuse of notation, so thanks for bringing that up

Astyx

My pleasure :p

Secret

[Random] Do the notion "inverse proof", "identity proof" make sense. Can we have a group made of proofs?

DHMO

@Secret what is the operation?

morbidCode

Can someone help me? So I found about this approximation of pi created by someone called john wallis. It's like this: pi/4 = 2*4*4*6*6*8*.../3*3*5*5*7*7*.... Here's the link.
https://mitpress.mit.edu/sicp/full-text/sicp/book/node22.html
Now, I'll have to create a program that approximates pi using this formula. I don't know anything about series like this, but I do know what the output should be. My question is if the input of my program is 4, would the formula become 2*4/3*3 or 2*4*4/3*3*5? I have a feeling the second one is invalid, but I'm not sure why. Thanks for your help!

DHMO

@morbidCode why would it be invalid?

how the input correlates with the output is up to you

as long as when the input goes to infnity, the output goes to pi

what happens when the input is "4" is just your definition

meaning you can proceed two steps at a time...

2
2/3
2*4/3
2*4/3*3
2*4*4/3*3
2*4*4/3*3*5

morbidCode

13:21

@DHMO thanks! Got it!

Secret

@DHMO I have no idea, I only know for such structure, the elements are logical propositions

DHMO

@morbidCode my pleasure

@Secret I don't think "logical propositions" means the same thing as "proofs"...

Secret

Well I don't know if there is a term for the various steps of a proof

DHMO

@Secret you just said it

Secret

but then, what kind of structure do they form, given a step, you have many possible choices to move forward or backward, and that only depends on which axioms, lemma and theorem is being used

DHMO

13:24

how does that matter?

Secret

I knew that people do talk about the notion of a shortest proof, but it seems nothing is said about whether proofing something itself form an algebraic structure

DHMO

you can parameterize each step by the axioms used...

Astyx

Proof theory

Proof theory is a major branch of mathematical logic that represents proofs as formal mathematical objects, facilitating their analysis by mathematical techniques. Proofs are typically presented as inductively-defined data structures such as plain lists, boxed lists, or trees, which are constructed according to the axioms and rules of inference of the logical system. As such, proof theory is syntactic in nature, in contrast to model theory, which is semantic in nature. Some of the major areas of proof theory include structural proof theory, ordinal analysis, provability logic, reverse mathematics...

DHMO

@Astyx right, grammar is also a part of mathematics lol

NaCl

I need to prove that the axiom of completeness is equivalent to the Least-upper-bound property of $\mathbb{R}$. fml

Astyx

13:31

:p

Secret

Reverse mathematics

Reverse mathematics is a program in mathematical logic that seeks to determine which axioms are required to prove theorems of mathematics. Its defining method can briefly be described as "going backwards from the theorems to the axioms", in contrast to the ordinary mathematical practice of deriving theorems from axioms. It can be conceptualized as sculpting out necessary conditions from sufficient ones. The reverse mathematics program was foreshadowed by results in set theory such as the classical theorem that the axiom of choice and Zorn's lemma are equivalent over ZF set theory. The goal of reverse...

Ah, so that is what I was unawaringly doing when trying to break axioms

->Optimisation of a reverse mathematics procedure

6

Q: When does something become a "mathematical object"?

A mathematical object is an abstract object arising in philosophy of mathematics and mathematics. Abstract object: Abstract and concrete are classifications that denote whether a term describes an object with a physical referent or one with no physical referents. Denoting, specifying, ...

Ok I once again drifted accidentally into metamathematics

Hiro

13:56

@DHMO Regarding yesterday's question, I just found that (11), (33), (55), (77), and (99) are the positive integers divisible by b+1 and under b^3. The quadratic ones don't work because either the sum is even or the product is even for every single one. Also if we do something like (b+1)(3b-1), that gives me 3b^2+2b-1 (now 2 is even and -1 doesn't make sense to put as a digit in the representation), do you agree, or rather is the conclusion I have drawn correct?

DHMO

@Hiro 3b^2 + b + (b-1)

e.g. when b=6, we have (315)

Hiro

How can you have b-1 in the units?

DHMO

because 0 <= (b-1) < b

any digit is allowed as long as 0 <= that digit < b

user379685

Hello, I have to approximate this approximation sinx = x-x^3/6, |x|<pi/6. Can I do it like this error(x)=sinx - x + x^3/6 and the approximation is the maximal value of error(x), |x|<pi/6

Hiro

so what does 3b^2 + b + (b-1) represent in terms of the number of positive integers? @DHMO

DHMO

13:58

@Hiro a counterexample of your claim that there exists no three-digit numbers satisfying your constraint

@user379685 are you approximating an approximation??

user379685

english isn't my first language, i have to evaluate how good the left formula is

right*

DHMO

alright, continue

user379685

there shouldn't be = but this ~

So can i do it like i described above?

DHMO

yes

Hiro

Why can't the digit be greater than b @DHMO ?

DHMO

14:02

@Hiro well that's the rule

if it is greater than b, you can subtract b and carry 1 to the left digit

Hiro

does b^2 + b + (b-1) work as well?

user379685

but there's also the taylor series and i get a different anwser using it's reminder

DHMO

it isn't divisible by b+1

@user379685 why?

user379685

error(x) = sin x - x + x^3/6, |x|<pi/6, it's maxmial value is slightly less than error(pi/6), because pi/6 isn't in the domain

the reminder in the taylor series is the fifth derivative of sin (c) /5! * x^5

cos(c)/5!*x^5 < 1/5!*(pi/6)^5

hello?

DHMO

lol, I have no idea

user379685

14:15

:C

Hiro

@DHMO (313),(315),(317),(319),(333),(335),(337),(339),(353),(355),(357),(359),(373),(3‌75),(377),(379),(393),(395),(397),(399) <-- All work correct?

DHMO

@Hiro of course not

Hiro

Why not? @DHMO

DHMO

well, because they are not divisible by b+1

Hiro

what about (333)?

DHMO

14:22

why would you think (333) is divisible by b+1?

Hiro

if b=4?

s.harp

you mean 2?

DHMO

@Hiro when b=4, (333) = 63 and b+1 =5.

Null

. xkcd.com/732

Hiro

Ah, so we have to check if the representation is divisible by b+1

DHMO

14:25

@Hiro I thought that is stated in the beginning.

Hiro

but how come (315) works?

DHMO

(315) only works when b=6.

Hiro

When I do (315)= 3b^2 + b +5

DHMO

As I said, it is 3b^2 + b + (b-1)

(31[b-1])

you can't represent [b-1] as a single digit, but it is still a valid digit in base-b.

Hiro

and is that the only that works, as in there's no (1,1,(b-1)) or something

DHMO

14:28

no that is not

that was meant to be a hint

Hiro

does b-2, b-3 ... etc work?

actually b-2 doesn't work

DHMO

try

Hiro

alright ill try and get back to u, one sec

NaCl

So.... $B:=\{f:f\text{ is a function from }\mathbb{Q}\to\mathbb{N}\}$ is countable isn't it? How the heck do I prove that? Yeah sure, by showing that there exists a bijective mapping, but.... A mapping from a set of functions to $\mathbb{N}$ is a bit unintuitive

DHMO

@NaCl functions are just sets.

A function from $\Bbb Q$ to $\Bbb N$ can be expressed as a subset of $\Bbb Q\times \Bbb N$

NaCl

14:40

Yeah, sure

Rajath Krishna R

I have a query. It is just to clarify something and I am not sure whether it could be put up as a question. Given a set of vectors v_1,v_2,....v_n are there infinite number of matrices with v_1,....v_n as eigenvectors?

DHMO

For example, $f(n)=2n$ can be represented by $\{(x,y)|y=2x\}$ (sorry for using the wrong domain and image)

@NaCl you can map the set bijectively to $\mathscr P(\Bbb N)$, which makes it uncountable

NaCl

I wish I could see that as fast and as clear as you

@DHMO for real, how did you find that?

DHMO

@NaCl I see a function as a set with cardinality equal to the domain

I see functions in terms of relations.

$f:S\to T \implies f\subseteq S \times T \land |f| = |S|$

NaCl

But $\mathbb{Q}\times\mathbb{N}$ is countable, isn't it?

DHMO

14:50

@NaCl yes

NaCl

Wait, I think I see it

Astyx

Formally a function is a triplet $(\Gamma, E, F)$ where $E$ is the domain, $F$ the codomain, and $\Gamma$ a graph on $E\times F$, ie $\Gamma \subset E\times F$ and $\forall x\in E, \exists ! y \in F, (x,y)\in \Gamma$

NaCl

All possible functions should correspond to all possible subsets of $\mathbb{Q}\times\mathbb{N}$, which is of size $2^\text{whatever}$

I wonder if that phrase is enough as "proof"

DHMO

yes

@Astyx thanks

Hiro

@DHMO I figured out a pattern! (3,1,b-1), (5,1,b-3), (7,1,b-5), (9,1,b-7) <-- these work correct?

DHMO

14:55

@Hiro nice

Hiro

what happens if I change the middle unit to 3?

DHMO

@Hiro try

NaCl

@DHMO Actually no, let me rephrase, shouldn't the subset be from $\mathbb{N}$, let's call all subsets of $\mathbb{N}$ $T$, such that all possible functions should corrsepons to $\mathbb{Q}\times T$ and not to all subsets of $\mathbb{Q}\times\mathbb{N}$?

DHMO

@NaCl I'm saying that each function is a subset of $\Bbb Q \times \Bbb N$

NaCl

Why? The set consists of all functions which are mapped from $\mathbb{Q}\to\mathbb{N}$

So $\mathbb{Q}$ must remain as it is, mustn't it?

DHMO

14:59

I mean the function themselves

NaCl

so my previous wording was acceptable?

heather

hello everyone

DHMO

@NaCl I think so....

Transcript for

$Mathematics$ Mathematics